Solution by David W. Carter, Draper Laboratory, Cambridge, MA. The minimum rank of the combination is Fr/21, where r = rank A. It is convenient to work in a more general setting. Let K be a field, A an m-by-n matrix over K, and ae a nonzero element of an algebraic extension of K. Define p(K, A, ao) = minB rank (A + aoB), where B ranges over m-by-n matrices with entries in K. If S is an invertible matrix of order m over K, then p(K, SA, a) = p(K, A, a). Similarly, p(K, AT, o!) = p(K, A, a!) when T is an invertible matrix of order n. By elementary linear algebra, there exist invertible matrices S and T such that SAT = E, where E = (Ir 0). Hence it suffices to determine p(K, E, a). Let B* be an m-by-n matrix such that E + aB* achieves the minimum rank. We may assume that the last m -r rows of B* are zero, since the dimension of the vector space over L spanned by the first r rows of E + aB* is not more than the dimension of the space spanned by all m rows. Similarly, we may assume that the rightmost n -r columns of B* are zero. Thus B* has the form (C O), where C is an r-by-r matrix over K. The rank of E + aoB* is the rank of 1r + oaC. This is less than r if and only if -a71l is an eigenvalue of C. Indeed, the rank of Ir + aoC equals r s, where s is the dimension of the eigenspace associated with eigenvalue -o-1 of C. Let f be an irreducible polynomial over K with -a-1 as a root. The degree d of f is the degree of the field extension [K(a): K]. If C has s linearly independent eigenvectors with eigenvalue -a-1, then fS divides the characteristic polynomial of C. Thus ds < r. Conversely, if s is a positive integer such that ds < r, then we construct an rby-r block diagonal matrix over K with s linearly independent eigenvectors having eigenvalue -o-1: use s blocks equal to the d-by-d companion matrix for f and one remaining block Ir-ds* We have shown that p(K,A,a) = Fr/dl. When K = IR and oa = i, we have d = 2.